Question

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10^5 ms–1 . If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.
Ncert solutions for Class 11th Chemistry Part - 1 Chapter 2 Exercise 60

Answer 1

potential difference, ∆V = 1000 volts

velocity of proton , v = 4.37 × 10^5 m/s

mass of hockey ball , m = 0.1 kg

velocity of hockey ball = velocity of proton = 4.37 × 10^5 m/s

we have to find wavelength associated with hockey ball.

using De-broglie's wavelength equation,

wavelength = h/mv

= 6.63 × 10^-34 Js/(0.1kg × 4.37 × 10^5m/s)

= 6.63 × 10^-34/(0.437 × 10^5)

= 15.17 × 10^-39 m

= 1.517 × 10^-38 m

hence, answer is 1.517 × 10^-38 m

Answer 2

Answer : The wavelength is $1.52\times 10^{-38}kg$

Explanation :

According to de-Broglie, the expression for wavelength is,

$\lambda=\frac{h}{p}$

and,

$p=mv$

where,  

p = momentum, m = mass, v = velocity

So, the formula will be:

$\lambda=\frac{h}{mv}$      .............(1)

where,

$\lambda$ = wavelength = ?

h = Planck's constant = $6.626\times 10^{-34}Js=6.626\times 10^{-34}kgm^2/s$

v = velocity = $4.37\times 10^5m/s$

m = mass = 0.1 kg

Now put all the given values in equation 1, we get:

$\lambda=\frac{6.626\times 10^{-34}kgm^2/s}{0.1kg\times (4.37\times 10^5m/s)}$

$\lambda=1.52\times 10^{-38}kg$

Therefore, the wavelength is $1.52\times 10^{-38}kg$